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Assignment 3 solutions

# Assignment 3 solutions - MIT OpenCourseWare...

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MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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18.085 - Mathematical Methods for Engineers I Prof. Gilbert Strang Solutions - Problem Set 3 Section 1.6 1 1 1) T = 1 2 ⎦ ⎥ u T Tu = u 1 u 2 1 1 u 1 1 2 u 2 u 1 u 2 = u 1 u 2 u 1 + 2 u 2 = u 1 2 u 1 u 2 + ( u 1 u 2 + 2 u 2 2 ) = u 1 2 2 u 1 u 2 + 2 u 2 2 = ( u 1 u 2 ) 2 + u 2 2 > 0 # u T Tu is positive definite as it is the sum of 2 squares 1 1 0 2 1 1 3) A = 0 1 1 A T A = 1 2 1 1 0 1 1 1 2 Au = 0 A T Au = 0 A is a singular matrix as it has only 2 linearly independent columns. Since A does not have full rank, A T A will have a zero pivot 2 1 1 A T A = 1 2 1 1 1 2 2 1 1 0 3 / 2 3 / 2 row 2 row 2 + 1 row 1 = 2 0 3 / 2 3 / 2 row 3 row 3 + 1 2 row 1 2 1 1 = 0 3 / 2 3 / 2 0 0 0 row 3 row 3 + row 2 A T A is only semidefinite # Au = 0, A T Au = 0 The nullvector of A and A T A would be a constant vector ⎡ � 1 1 0 c 0 0 1 1 ⎣ � c = 0 1 0 1 c 0 ⎡ � 2 1 1 c 0 1 2 1 ⎣ � c = 0 1 1 2 c 0 1
c u = c where c is a constant # c 1 b 6) K = b 4 ⎦ ⎥ u T Ku = u 1 u 2 1 b u 1 b 4 u 2 u 1 + bu 2 = u 1 u 2 bu 1 + 4 u 2 = u 1 2 + bu 1 u 2 + bu 1 u 2 + 4 u 2 2 = u 1 2 + 2 bu 1 u 2 + 4 u 2 2 = ( u 1 + bu 2 ) 2 + (4 b 2 ) u 2 2 > 0 4 b 2 > 0 ( b + 2)( b 2) < 0 2 2 2 < b < 2 # For semidefinite case, the borderline value of b is 2 and 2 u T Ku = ( u 1 + bu 2 ) 2 (only one square) 1 5 If b = 5, K = 5 4 By Gaussian Elimination 1 5 5 4 1 5 = 0 21 row 2 row 2 5 × row 1 The pivots are 1 and 21 The matrix is indefinite if b = 5 # 11) f ( x, y ) = 2 xy �f �f = 2 y = 2 x �x �y 2 f 2 f 2 f = 2 , = 0 , = 0 �x�y �x 2 �y 2 2

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0 2 Hessian matrix, H = 2 0 a b x x y 2 xy b c y ax + by = x y bx + cy = ax 2 + bxy + bxy + cy 2 = ax 2 + 2 bxy + cy 2 By comparing coeﬃcients, a = 0, c = 0 b = 1 0 1 The symmetric matrix that produces f ( x, y ) = 2 xy is S = 1 0 det( S �I ) = 0 1 det = 0 1 2 1 = 0 ( 1)( + 1) = 0 = 1 or 1 The
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